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Proposition 10.5. \( \widehat{G} \cong \widehat{G} \) . | ∎ | No |
Lemma 10.6. If \( \left( {M}_{n}\right) ,\left( {M}_{n}^{\prime }\right) \) are stable \( \mathfrak{a} \) -filtrations of \( M \), then they have bounded difference: that is, there exists an integer \( {n}_{0} \) such that \( {M}_{n + {n}_{0}} \subseteq {M}_{n}^{\prime } \) and \( {M}_{n + {n}_{0}}^{\prime } \subseteq ... | Proof. Enough to take \( {M}_{n}^{\prime } = {\mathfrak{a}}^{n}M \) . Since \( \mathfrak{a}{M}_{n} \subseteq {M}_{n + 1} \) for all \( n \), we have \( {\mathfrak{a}}^{n}M \subseteq {M}_{n} \) ; also \( \mathfrak{a}{M}_{n} = {M}_{n + 1} \) for all \( n \geq {n}_{0} \) say, hence \( {M}_{n + {n}_{0}} = {\mathfrak{a}}^{n... | Yes |
Proposition 10.7. The following are equivalent, for a graded ring \( A \) :\ni) \( A \) is a Noetherian ring;\nii) \( {A}_{0} \) is Noetherian and \( A \) is finitely generated as an \( {A}_{0} \) -algebra. | Proof. i) \( \Rightarrow \) ii). \( {A}_{0} \cong A/{A}_{ + } \), hence is Noetherian. \( {A}_{ + } \) is an ideal in \( A \), hence is finitely generated, say by \( {x}_{1},\ldots ,{x}_{s} \), which we may take to be homogeneous elements of \( A \), of degrees \( {k}_{1},\ldots ,{k}_{s} \) say (all \( > 0 \) ). Let \(... | Yes |
Lemma 10.8. Let \( A \) be a Noetherian ring, \( M \) a finitely-generated \( A \) -module,\n\n\( \left( {M}_{n}\right) \) an a-filtration of \( M \) . Then the following are equivalent:\n\n i) \( {M}^{ * } \) is a finitely-generated \( {A}^{ * } \) -module;\n\n ii) The filtration \( \left( {M}_{n}\right) \) is stable. | Proof. Each \( {M}_{n} \) is finitely generated, hence so is each \( {Q}_{n} = {\bigoplus }_{r = 0}^{n}{M}_{r} \) : this is a subgroup of \( {M}^{ * } \) but not (in general) an \( {A}^{ * } \) -submodule. However, it generates one, namely\n\n\[ \n{M}_{n}^{ * } = {M}_{0} \oplus \cdots \oplus {M}_{n} \oplus \mathfrak{a}... | Yes |
Proposition 10.9. (Artin-Rees lemma). Let \( A \) be a Noetherian ring, a an ideal in \( A, M \) a finitely-generated \( A \) -module, \( \left( {M}_{n}\right) \) a stable a-filtration of \( M \) . If \( {M}^{\prime } \) is a submodule of \( M \), then \( \left( {{M}^{\prime } \cap {M}_{n}}\right) \) is a stable a-filt... | Proof. We have \( \mathfrak{a}\left( {{M}^{\prime } \cap {M}_{n}}\right) \subseteq \mathfrak{a}{M}^{\prime } \cap \mathfrak{a}{M}_{n} \subseteq {M}^{\prime } \cap {M}_{n + 1} \), hence \( \left( {{M}^{\prime } \cap {M}_{n}}\right) \) is an a-filtration. Hence it defines a graded \( {A}^{ * } \) -module which is a submo... | Yes |
Proposition 10.13. For any ring \( A \), if \( M \) is finitely-generated, \( \widehat{A}{ \otimes }_{A}M \rightarrow \widehat{M} \) is surjective. If, moreover, \( A \) is Noetherian then \( \widehat{A}{ \otimes }_{A}M \rightarrow \widehat{M} \) is an isomorphism. | Proof. Using (10.3) or otherwise it is clear that a-adic completion commutes with finite direct sums. Hence if \( F \cong {A}^{n} \) we have \( \widehat{A}{ \otimes }_{A}F \cong \widehat{F} \) . Now assume \( M \) is finitely generated so that we have an exact sequence\n\n\[ 0 \rightarrow N \rightarrow F \rightarrow M ... | Yes |
Proposition 10.15. If \( A \) is Noetherian, \( \widehat{A} \) its \( \mathfrak{a} \) -adic completion, then\n\ni) \( \widehat{\mathfrak{a}} = \widehat{A}\mathfrak{a} \cong \widehat{A}{ \otimes }_{A}\mathfrak{a} \) ;\n\nii) \( {\left( {\mathfrak{a}}^{n}\right) }^{ \frown } = {\left( \widehat{\mathfrak{a}}\right) }^{n} ... | Proof. Since \( A \) is Noetherian, \( \mathfrak{a} \) is finitely-generated. (10.13) implies that the map\n\n\[ \widehat{A}{ \otimes }_{A}\mathfrak{a} \rightarrow \widehat{\mathfrak{a}} \]\n\nwhose image is \( \widehat{A}a \), is an isomorphism. This proves i). Now apply i) to \( {a}^{n} \) and we deduce that\n\n\[ {\... | Yes |
Proposition 10.16. Let \( A \) be a Noetherian local ring, \( \mathfrak{m} \) its maximal ideal. Then the \( \mathfrak{m} \) -adic completion \( \widehat{A} \) of \( A \) is a local ring with maximal ideal \( \widehat{\mathfrak{m}} \) . | Proof. By (10.15) iii) we have \( \widehat{A}/\widehat{\mathfrak{m}} \cong A/\mathfrak{m} \), hence \( \widehat{A}/\widehat{\mathfrak{m}} \) is a field and so \( \widehat{\mathfrak{m}} \) is a maximal ideal. By (10.15) iv) it follows that \( \widehat{m} \) is the Jacobson radical of \( \widehat{A} \) and so is the uniq... | No |
Theorem 10.17. Let \( A \) be a Noetherian ring, \( \mathfrak{a} \) an ideal, \( M \) a finitely-generated \( A \) -module and \( \widehat{M} \) the \( \mathfrak{a} \) -completion of \( M \) . Then the kernel \( E = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathfrak{a}}^{n}M \) of \( M \rightarrow \widehat{M} \) co... | Proof. Since \( E \) is the intersection of all neighborhoods of \( 0 \in M \), the topology induced on it is trivial, i.e., \( E \) is the only neighborhood of \( 0 \in E \) . By (10.11) the induced topology on \( E \) coincides with its \( \mathfrak{a} \) -topology. Since \( \mathfrak{a}E \) is a neighborhood in the ... | Yes |
Corollary 10.18. Let \( A \) be a Noetherian domain, \( \mathfrak{a} \neq \left( 1\right) \) an ideal of \( A \) . Then \( \bigcap {\mathfrak{a}}^{n} = 0 \) . | Proof. \( 1 + \alpha \) contains no zero-divisors. | No |
Corollary 10.19. Let \( A \) be a Noetherian ring, \( \mathfrak{a} \) an ideal of \( A \) contained in the Jacobson radical and let \( M \) be a finitely-generated \( A \) -module. Then the \( \mathfrak{a} \) -topology of \( M \) is Hausdorff, i.e. \( ! \cap {\mathfrak{a}}^{n}M = 0 \) . | Proof. By (1.9) every element of \( 1 + \alpha \) is a unit. | No |
Proposition 10.22. Let \( A \) be a Noetherian ring, \( \mathfrak{a} \) an ideal of \( A \). Then\ni) \( {G}_{\mathfrak{a}}\left( A\right) \) is Noetherian;\nii) \( {G}_{\mathfrak{a}}\left( A\right) \) and \( {G}_{\widehat{\mathfrak{a}}}\left( \widehat{A}\right) \) are isomorphic as graded rings;\niii) if \( M \) is a ... | Proof. i) Since \( A \) is Noetherian, \( \alpha \) is finitely generated, say by \( {x}_{1},\ldots ,{x}_{s} \). Let \( {\bar{x}}_{i} \) be the image of \( {x}_{i} \) in \( \mathfrak{a}/{\mathfrak{a}}^{2} \), then \( G\left( A\right) = \left( {A/\mathfrak{a}}\right) \left\lbrack {{\bar{x}}_{1},\ldots ,{\bar{x}}_{s}}\ri... | Yes |
Lemma 10.23. Let \( \phi : A \rightarrow B \) be a homomorphism of filtered groups, i.e. \( \phi \left( {A}_{n}\right) \subseteq {B}_{n} \), and let \( G\left( \phi \right) : G\left( A\right) \rightarrow G\left( B\right) ,\widehat{\phi } : \widehat{A} \rightarrow \widehat{B} \) be the induced homomorphisms of the assoc... | Proof. Consider the commutative diagram of exact sequences\n\n\[ 0 \rightarrow {A}_{n}/{A}_{n + 1} \rightarrow A/{A}_{n + 1} \rightarrow A/{A}_{n} \rightarrow 0 \]\n\n\[ { \downarrow }^{{G}_{n}\left( \phi \right) }\;{ \downarrow }^{{\alpha }_{n + 1}}\;{ \downarrow }^{{\alpha }_{n}} \]\n\n\[ 0 \rightarrow {B}_{n}/{B}_{n... | Yes |
Proposition 10.24. Let \( A \) be a ring, \( \mathfrak{a} \) an ideal of \( A, M \) an \( A \) -module, \( \left( {M}_{n}\right) \) an \( \mathfrak{a} \) -filtration of \( M \) . Suppose that \( A \) is complete in the \( \mathfrak{a} \) -topology and that \( M \) is Hausdorff in its filtration topology (i.e. that \( \... | Proof. Pick a finite set of generators of \( G\left( M\right) \), and split them up into their homogeneous components, say \( {\xi }_{i}\left( {1 \leq i \leq \nu }\right) \) where \( {\xi }_{i} \) has degree say \( n\left( i\right) \), and is therefore the image of say \( {x}_{i} \in {M}_{n\left( i\right) } \) . Let \(... | Yes |
Corollary 10.25. With the hypotheses of (10.24), if \( G\left( M\right) \) is a Noetherian \( G\left( A\right) \) -module, then \( M \) is a Noetherian \( A \) -module. | Proof. We have to show that every submodule \( {M}^{\prime } \) of \( M \) is finitely generated (6.2). Let \( {M}_{n}^{\prime } = {M}^{\prime } \cap {M}_{n} \) ; then \( \left( {M}_{n}^{\prime }\right) \) is an \( \mathfrak{a} \) -filtration of \( {M}^{\prime } \), and the embedding \( {M}_{n}^{\prime } \rightarrow {M... | Yes |
Theorem 10.26. If \( A \) is a Noetherian ring, \( \mathfrak{a} \) an ideal of \( A \), then the \( \mathfrak{a} \) - completion \( \widehat{A} \) of \( A \) is Noetherian. | Proof. By (10.22) we know that\n\n\[ \n{G}_{\mathfrak{a}}\left( A\right) = {G}_{\mathfrak{d}}\left( \widehat{A}\right) \n\]\n\nis Noetherian. Now apply (10.25) to the complete ring \( \widehat{A} \), taking \( M = \widehat{A} \) (filtered by \( {\widehat{\mathfrak{a}}}^{n} \), and so Hausdorff). | No |
Corollary 10.27. If \( A \) is a Noetherian ring, the power series ring \( B = \) \( A\left\lbrack \left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \right\rbrack \) in \( n \) variables is Noetherian. In particular \( k\left\lbrack \left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \right\rbrack \) (k a field) is N... | Proof. \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is Noetherian by the Hilbert basis theorem, and \( B \) is its completion for the \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) -adic topology. | Yes |
Theorem 11.1. (Hilbert, Serre). \( P\left( {M, t}\right) \) is a rational function in \( t \) of the form \( f\left( t\right) /\mathop{\prod }\limits_{{i = 1}}^{s}\left( {1 - {t}^{{k}_{i}}}\right) \), where \( f\left( t\right) \in \mathbf{Z}\left\lbrack t\right\rbrack \) . | Proof. By induction on \( s \), the number of generators of \( A \) over \( {A}_{0} \) . Start with \( s = 0 \) ; this means that \( {A}_{n} = 0 \) for all \( n > 0 \), so that \( A = {A}_{0} \) and \( M \) is a finitely-generated \( {A}_{0} \) module, hence \( {M}_{n} = 0 \) for all large \( n \) . Thus \( P\left( {M,... | Yes |
Corollary 11.2. If each \( {k}_{i} = 1 \), then for all sufficiently large \( n,\lambda \left( {M}_{n}\right) \) is a polynomial in \( n \) (with rational coefficients) of degree* \( d - 1 \) . | Proof. By (11.1) we have \( \lambda \left( {M}_{n}\right) = \) coefficient of \( {t}^{n} \) in \( f\left( t\right) \cdot {\left( 1 - t\right) }^{-s} \) . Canceling powers of \( \left( {1 - t}\right) \) we may assume \( s = d \) and \( f\left( 1\right) \neq 0 \) . Suppose \( f\left( t\right) = \) \( \mathop{\sum }\limit... | Yes |
Proposition 11.3. If \( x \in {A}_{k} \) is not a zero-divisor in \( M \) then \( d\left( {M/{xM}}\right) = \)\n\n\( d\left( M\right) - 1 \) . | We shall use (11.1) in the case where \( {A}_{0} \) is an Artin ring (in particular, a field) and \( \lambda \left( M\right) \) is the length \( l\left( M\right) \) of a finitely-generated \( {A}_{0} \) -module \( M \) . By (6.9) \( l\left( M\right) \) is additive. | No |
Proposition 11.4. Let \( A \) be a Noetherian local ring, \( \mathrm{m} \) its maximal ideal, \( \mathfrak{q} \) an \( \mathfrak{m} \) -primary ideal, \( M \) a finitely-generated \( A \) -module, \( \left( {M}_{n}\right) \) a stable \( \mathfrak{q} \) - filtration of \( M \) . Then\ni) \( M/{M}_{n} \) is of finite len... | Proof. i) Let \( G\left( A\right) = {\bigoplus }_{n}{\mathfrak{q}}^{n}/{\mathfrak{q}}^{n + 1}, G\left( M\right) = {\bigoplus }_{n}{M}_{n}/{M}_{n + 1}.{G}_{0}\left( A\right) = A/\mathfrak{q} \) is an Artin local ring, say by (8.5); \( G\left( A\right) \) is Noetherian, and \( G\left( M\right) \) is a finitely-generated ... | Yes |
Proposition 11.6. If \( A,\mathfrak{m},\mathfrak{q} \) are as above\n\n\[ \deg {\chi }_{\mathfrak{q}}\left( n\right) = \deg {\chi }_{\mathfrak{m}}\left( n\right) \] | Proof. We have \( \mathfrak{m} \supseteq \mathfrak{q} \supseteq {\mathfrak{m}}^{r} \) for some \( r \) by (7.16), hence \( {\mathfrak{m}}^{n} \supseteq {\mathfrak{q}}^{n} \supseteq {\mathfrak{m}}^{rn} \) and therefore\n\n\[ {\chi }_{\mathfrak{m}}\left( n\right) \leq {\chi }_{\mathfrak{q}}\left( n\right) \leq {\chi }_{\... | Yes |
Proposition 11.8. Let \( A,\mathfrak{m},\mathfrak{q} \) be as before. Let \( M \) be a finitely-generated \( A \) -module, \( x \in A \) a non-zero-divisor in \( M \) and \( {M}^{\prime } = M/{xM} \) . Then\n\n\[\n\deg {\chi }_{q}^{{M}^{\prime }} \leq \deg {\chi }_{q}^{M} - 1\n\] | Proof. Let \( N = {xM} \) ; then \( N \cong M \) as \( A \) -modules, by virtue of the assumption on \( x \) . Let \( {N}_{n} = N \cap {\mathfrak{q}}^{n}M \) . Then we have exact sequences\n\n\[0 \rightarrow N/{N}_{n} \rightarrow M/{\mathfrak{q}}^{n}M \rightarrow {M}^{\prime }/{\mathfrak{q}}^{n}{M}^{\prime } \rightarro... | Yes |
Corollary 11.9. If \( A \) is a Noetherian local ring, \( x \) a non-zero-divisor in \( A \), then \( d\left( {A/\left( x\right) }\right) \leq d\left( A\right) - 1 \) . | Proof. Put \( M = A \) in (11.8). | No |
Proposition 11.10. \( d\left( A\right) \geq \dim A \) . | Proof. By induction on \( d = d\left( A\right) \) . If \( d = 0 \) then \( l\left( {A/{\mathfrak{m}}^{n}}\right) \) is constant for all large \( n \), hence \( {\mathfrak{m}}^{n} = {\mathfrak{m}}^{n + 1} \) for some \( n \), hence \( {\mathfrak{m}}^{n} = 0 \) by Nakayama’s lemma (2.6). Thus \( A \) is an Artin ring and... | Yes |
If \( A \) is a Noetherian local ring, \( \dim A \) is finite. | ∎ | No |
Proposition 11.13. Let \( A \) be a Noetherian local ring of dimension \( d \) . Then there exists an \( \mathfrak{m} \) -primary ideal in \( A \) generated by \( d \) elements \( {x}_{1},\ldots ,{x}_{a} \) , and therefore \( \dim A \geq \delta \left( A\right) \) . | Proof. Construct \( {x}_{1},\ldots ,{x}_{d} \) inductively in such a way that every prime ideal containing \( \left( {{x}_{1},\ldots ,{x}_{i}}\right) \) has height \( \geq i \), for each \( i \) . Suppose \( i > 0 \) and \( {x}_{1},\ldots ,{x}_{i - 1} \) constructed. Let \( {\mathfrak{p}}_{j}\left( {1 \leq j \leq s}\ri... | Yes |
Theorem 11.14. (Dimension theorem.) For any Noetherian local ring \( A \) the following three integers are equal:\n\ni) the maximum length of chains of prime ideals in \( A \) ;\n\nii) the degree of the characteristic polynomial \( {\chi }_{\mathfrak{m}}\left( n\right) = l\left( {A/{\mathfrak{m}}^{n}}\right) \) ;\n\nii... | Proof. (11.7), (11.10), (11.13). | No |
Corollary 11.15. \( \dim A \leq {\dim }_{k}\left( {\mathfrak{m}/{\mathfrak{m}}^{2}}\right) \) . | Proof. If \( {x}_{i} \in \mathfrak{m}\left( {1 \leq i \leq s}\right) \) are such that their images in \( \mathfrak{m}/{\mathfrak{m}}^{2} \) form a basis of this vector space, then the \( {x}_{i} \) generate \( m \) by \( \left( {2.8}\right) \) ; hence \( {\dim }_{k}\left( {m/{m}^{2}}\right) = \) \( s \geq \dim A \) by ... | Yes |
Corollary 11.16. Let \( A \) be a Noetherian ring, \( {x}_{1},\ldots ,{x}_{r} \in A \) . Then every minimal ideal \( \mathfrak{p} \) belonging to \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) has height \( \leq r \) . | Proof. In \( {A}_{\mathfrak{p}} \) the ideal \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) becomes \( {\mathfrak{p}}^{e} \) -primary, hence \( r \geq \dim {A}_{\mathfrak{p}} = \) height \( \mathfrak{p} \) . | Yes |
Corollary 11.17. (Krull's principal ideal theorem). Let \( A \) be a Noetherian ring and let \( x \) be an element of \( A \) which is neither a zero-divisor nor a unit. Then every minimal prime ideal \( \mathfrak{p} \) of \( \left( x\right) \) has height 1. | Proof. By (11.16), height \( \mathfrak{p} \leq 1 \) . If height \( \mathfrak{p} = 0 \), then \( \mathfrak{p} \) is a prime ideal belonging to 0, hence every element of \( \mathfrak{p} \) is a zero-divisor by (4.7): contradiction, since \( x \in \mathfrak{p} \) . | Yes |
Corollary 11.18. Let \( A \) be a Noetherian local ring, \( x \) an element of \( \mathfrak{m} \) which is not a zero-divisor. Then \( \dim A/\left( x\right) = \dim A - 1 \) . | Proof. Let \( d = \dim A/\left( x\right) \) . By (11.9) and (11.14) we have \( d \leq \dim A - 1 \) . On the other hand, let \( {x}_{i}\left( {1 \leq i \leq d}\right) \) be elements of \( m \) whose images in \( A/\left( x\right) \) generate an \( \mathfrak{m}/\left( x\right) \) -primary ideal. Then the ideal \( \left(... | Yes |
Corollary 11.19. Let \( \widehat{A} \) be the \( \mathfrak{m} \) -adic completion of \( A \) . Then \( \dim A = \) \( \dim \widehat{A} \) . | Proof. \( A/{\mathfrak{m}}^{n} \cong \widehat{A}/{\widehat{\mathfrak{m}}}^{n} \) from (10.15), hence \( {\chi }_{\mathfrak{m}}\left( n\right) = {\chi }_{\widehat{\mathfrak{m}}}\left( n\right) \) . ∎ | Yes |
Proposition 11.20. Let \( {x}_{1},\ldots ,{x}_{d} \) be a system of parameters for \( A \) and let \( \mathfrak{q} = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \) be the \( \mathfrak{m} \) -primary ideal generated by them. Let \( f\left( {{t}_{1},\ldots ,{t}_{d}}\right) \) be a homogeneous polynomial of degree \( s \) wit... | Proof. Consider the epimorphism of graded rings\n\n\[ \alpha : \left( {A/\mathfrak{q}}\right) \left\lbrack {{t}_{1},\ldots ,{t}_{d}}\right\rbrack \rightarrow {G}_{\mathfrak{q}}\left( A\right) \]\n\ngiven by \( {t}_{i} \rightarrow {\bar{x}}_{i} \), where \( {t}_{i} \) are indeterminates and \( {\bar{x}}_{i} \) is \( {x}... | Yes |
Corollary 11.21. If \( k \subset A \) is a field mapping isomorphically onto \( A/\mathfrak{m} \) and if \( {x}_{1},\ldots ,{x}_{d} \) is a system of parameters, then \( {x}_{1},\ldots ,{x}_{d} \) are algebraically independent over \( k \) . | Proof. Assume \( f\left( {{x}_{1},\ldots ,{x}_{d}}\right) = 0 \) where \( f \) is a polynomial with coefficients in \( k \) . If \( f ≢ 0 \) we can write \( f = {f}_{s} + \) higher terms, where \( {f}_{s} \) is homogeneous of degree \( s \) and \( {f}_{s} ≢ 0 \) . Apply (11.20) to \( {f}_{s} \) and we deduce that \( {f... | Yes |
Theorem 11.22. Let \( A \) be a Noetherian local ring of dimension \( d,\mathrm{\;m} \) its maximal ideal, \( k = A/\mathfrak{m} \) . Then the following are equivalent:\n\ni) \( {G}_{\mathfrak{m}}\left( A\right) \cong k\left\lbrack {{t}_{1},\ldots ,{t}_{d}}\right\rbrack \) where the \( {t}_{i} \) are independent indete... | Proof. i) \( \Rightarrow \) ii) is clear. ii) \( \Rightarrow \) iii) by (2.8): see the proof of (11.15). iii) \( \Rightarrow \) i): let \( \mathfrak{m} = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \), then by (11.20) the map \( \alpha : k\left\lbrack {{x}_{1},\ldots ,{x}_{d}}\right\rbrack \rightarrow {G}_{\mathfrak{m}}\le... | Yes |
Lemma 11.23. Let \( A \) be a ring, \( \mathfrak{a} \) an ideal \( \cdot \) of \( A \) such that \( \mathop{\bigcap }\limits_{n}{\mathfrak{a}}^{n} = 0 \). Suppose that \( {G}_{\mathfrak{a}}\left( A\right) \) is an integral domain. Then \( A \) is an integral domain. | Proof. Let \( x, y \) be non-zero elements of \( A \). Then since \( \cap {\mathfrak{a}}^{n} = 0 \) there exist integers \( r, s \geq 0 \) such that \( x \in {\mathfrak{a}}^{r}, x \notin {\mathfrak{a}}^{r + 1}, y \in {\mathfrak{a}}^{s}, y \notin {\mathfrak{a}}^{s + 1} \). Let \( \bar{x},\bar{y} \) denote the images of ... | Yes |
Proposition 11.24. Let \( A \) be a Noetherian local ring. Then \( A \) is regular if and only if \( \widehat{A} \) is regular. | Proof. By (10.16),(10.26) and (11.19) we know that \( \widehat{A} \) is a Noetherian local ring of the same dimension as \( A \) and with \( \widehat{m} \) as maximal ideal. Now use (10.22) which asserts that \( {G}_{\mathfrak{m}}\left( A\right) \cong {G}_{\mathfrak{m}}\left( \widehat{A}\right) \) and the result follow... | No |
Theorem 11.25. For any irreducible variety \( V \) over \( k \) the local dimension of \( V \) at any point is equal to \( \dim V \). | Remark. We already know by (11.21) that \( \dim V \geq \dim {A}_{\mathfrak{m}} \) for all \( \mathfrak{m} \) . The problem is to prove the opposite inequality, and for this purpose the main lemma is: | No |
Lemma 11.26. Let \( B \subseteq A \) be integral domains with \( B \) integrally closed and \( A \) integral over \( B \). Let \( \mathfrak{m} \) be a maximal ideal of \( A \), and let \( \mathfrak{n} = \mathfrak{m} \cap B \). Then \( \mathfrak{n} \) is maximal and \( \dim {A}_{\mathfrak{m}} = \dim {B}_{\mathfrak{n}} \... | Proof. This is an easy consequence of the results of Chapter 5. First \( \mathfrak{n} \) is maximal by (5.8). Next if \[ m \supset {q}_{1} \supset {q}_{2} \supset \cdots \supset {q}_{d} \] (1) is a strict chain of primes in \( A \), its intersection with \( B \) is by (5.9) a strict chain of primes \[ \mathfrak{n} \sup... | Yes |
Corollary 11.27. For every maximal ideal \( \mathfrak{m} \) of \( A\left( V\right) \) we have\n\n\[ \dim A\left( V\right) = \dim A{\left( V\right) }_{\mathfrak{m}}. \] | Proof. By definition we have \( \dim A\left( V\right) = \mathop{\sup }\limits_{\mathfrak{m}}\dim A{\left( V\right) }_{\mathfrak{m}} \) . But by (11.25) all \( A{\left( V\right) }_{\mathfrak{m}} \) have the same dimension. | Yes |
a. A \( \sqcup \) -semilattice satisfies the \( \sqcup \) -semilattice axioms, and \( x \leq y \) iff \( x \sqcup y = y \) . | Proof: For part a, certainly \( x \geq x \) and if \( u \geq x \) then \( u \geq x \), so \( x = x \sqcup x \) . The conditions \( w \geq x, y \) and whenever \( u \geq x,{yu} \geq w \) equally well specify \( x \sqcup y \) and \( y \sqcup x \) . Now, the following are equivalent:\n\n- \( x \sqcup \left( {y \sqcup z}\r... | Yes |
Corollary 8 (Lagrange’s theorem). Suppose \( G \) is a finite group. If \( H \subseteq G \) is a subgroup, then \( \left| H\right| \left| \right| G \mid \) . and for \( x \in {Go}\left( x\right) \left| \right| G \mid \) . | Proof: These are immediate consequences of the theorem. | No |
Lemma 10. Suppose \( h : G \mapsto {G}^{\prime } \) is a homomorphism, and \( S, T \subseteq G \) ; then \( h\left\lbrack {ST}\right\rbrack = h\left\lbrack S\right\rbrack h\left\lbrack T\right\rbrack \) . | Proof: We have \( x \in h\left\lbrack {ST}\right\rbrack \) iff \( x = h\left( {st}\right) \) for some \( s \in S, t \in T \), iff \( x = h\left( s\right) h\left( t\right) \) for some \( s \in S, t \in T \), iff \( x \in h\left\lbrack S\right\rbrack h\left\lbrack T\right\rbrack . | Yes |
Theorem 1 (Cayley’s theorem). A group is isomorphic to the group of its left translations. | Proof: The map \( a \mapsto {\psi }_{a} \) has already been shown to be an injective homomorphism; it is surjective since the group of left translations is defined to be the set of the \( {\psi }_{a} \) . | Yes |
Lemma 9. Units \( \left( {{R}_{1} \times \cdots \times {R}_{k}}\right) \) is isomorphic to Units \( \left( {R}_{1}\right) \times \cdots \times \operatorname{Units}\left( {R}_{k}\right) \) . | Proof: Clearly, \( \left\langle {{a}_{1},\ldots ,{a}_{n}}\right\rangle \) is a unit iff each \( {a}_{i} \) is. The isomorphism simply views Units \( \left( {R}_{1}\right) \times \cdots \times \) \( \operatorname{Units}\left( {R}_{k}\right) \) as a substructure of \( {R}_{1} \times \cdots \times {R}_{k} \) . | Yes |
Lemma 10 (Möbius inversion). For \( f, g \in A, x, y \in S, x \leq y \) , \[ g\left( {x, y}\right) = \mathop{\sum }\limits_{{w : x \leq w \leq y}}f\left( {w, y}\right) \text{ iff }f\left( {x, y}\right) = \mathop{\sum }\limits_{{w : x \leq w \leq y}}\mu \left( {x, w}\right) g\left( {w, y}\right) ;\text{ and } \] \[ g\le... | Proof: The first equation is equivalent to \( g = \chi * f \), the second to \( f = \mu * g \), the third to \( g = f * \chi \), and the fourth to \( f = g * \mu \) . The lemma follows by \( \mu = {\chi }^{-1} \) and algebra. | Yes |
Lemma 6.\n\n\[ \n{f}^{\left\lbrack i\right\rbrack }\left( {{x}_{0},\ldots ,{x}_{i}}\right) = \mathop{\sum }\limits_{{j = 0}}^{i}\frac{f\left( {x}_{j}\right) }{\mathop{\prod }\limits_{{k \neq j}}\left( {{x}_{j} - {x}_{k}}\right) }.\n\] | Proof: This may be proved by induction on \( i \) by a straightforward calculation. | No |
Lemma 8. Suppose \( R \) is an integral domain. If \( M \) is an \( R \) -module then \( M/\operatorname{Tor}\left( M\right) \) is torsion-free. | Proof: Let \( h \) be the canonical epimorphism and suppose \( {ah}\left( x\right) = 0, a \in R, a \neq 0 \) ; then \( h\left( {ax}\right) = 0 \), so \( {ax} \in \operatorname{Tor}\left( M\right) \), so \( {bax} = 0 \) for some \( b \in R \), so \( x \in \operatorname{Tor}\left( M\right) \), so \( h\left( x\right) = 0 ... | Yes |
Corollary 12. If \( M \) is a finitely generated torsion module over a principal ideal domain \( R \), then \( M \) is the direct sum of principal submodules \( {M}_{ij},1 \leq i \leq s,1 \leq j \leq {r}_{s} \) . The period of \( {M}_{ij} \) is \( {p}_{i}^{{e}_{ij}} \) for some prime element \( {p}_{i} \) and positive ... | Proof: Suppose \( {aM} = 0, a \neq 0 \) ; \( a \) has a factorization \( {p}_{1}^{{e}_{1}}\cdots {p}_{k}^{{e}_{k}} \) where \( {p}_{i} \) is prime element. By theorem \( {11M} \) is the direct sum of the \( {M}_{i} = {M}_{{p}_{i}^{{e}_{i}}} \) . Further \( {p}_{i}{M}_{i} = 0 \), and \( {M}_{i} \) is finitely generated ... | Yes |
Corollary 13. A finite commutative group is the direct sum of cyclic subgroups of prime power order; the orders of the subgroups are uniquely determined. | Proof: This follows immediately, since a finite commutative group is the same thing as a finitely generated torsion module over \( \mathcal{Z} \), and the submodules are the subgroups. | No |
Lemma 2. If \( E \supseteq F \) is finite and separable there is a \( c \in E \) such that \( E = F\left\lbrack c\right\rbrack \) . | Proof: It suffices to show the lemma for \( E = F\left\lbrack {a, b}\right\rbrack \), since then we may apply induction. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the embeddings over \( F \) of \( F\left\lbrack {a, b}\right\rbrack \) into its normal closure. If we can find a \( c \in F\left\lbrack {a, b}\right\r... | No |
Corollary 3. If \( E \supseteq F \) is separable and its elements all have degree \( \leq n \) then \( E \supseteq F \) is finite and \( \left\lbrack {E : F}\right\rbrack \leq n \) . | Proof: If not, choose \( L \) with \( F \subseteq L \subseteq E \) and and \( \left\lbrack {L : F}\right\rbrack > n \), and let \( c \) be such that \( L = F\left\lbrack c\right\rbrack \) ; then \( c \) has degree greater than \( n \), contradicting the hypothesis. | Yes |
Theorem 4 (Fundamental theorem of Galois theory). Suppose \( K \supseteq F \) is a Galois extension, and consider the maps \( E \mapsto {\operatorname{Aut}}_{E}\left( K\right) \) and \( G \mapsto \operatorname{Fixed}\left( G\right) \) between the fields \( E \) with \( F \subseteq E \subseteq K \) and the subgroups \( ... | Proof: Parts a and b are left to the reader. Part a states that the two maps form a Galois adjunction (see chapter 13), and part \( \mathrm{b} \) then follows by partial order theory. For part \( \mathrm{c} \), by remarks above \( K \supseteq E \) is Galois so if \( \alpha \notin E \) then \( \alpha \) is moved by some... | No |
Corollary 6. The multiplicative group of a finite field \( {\mathcal{F}}_{q} \) is cyclic. | The corollary also has various other proofs. For example, observe that, for \( d \mid q - 1 \) there are exactly \( d \) elements \( x \) of \( {\mathcal{F}}_{q}^{ \neq } \) such that \( {x}^{d} - 1 = 0 \) . This follows because \( {x}^{q - 1} - 1 \) has \( q - 1 \) roots, \( {x}^{d} - 1 \mid {x}^{q - 1} - 1 \) (exerci... | No |
Lemma 2. If \( {W}_{1},{W}_{2} \) are subspaces of a finite dimensional vector space \( V \) then \( \dim \left( {{W}_{1} + {W}_{2}}\right) = \dim \left( {W}_{1}\right) + \dim \left( {W}_{2}\right) - \dim \left( {{W}_{1} \cap {W}_{2}}\right) . | Proof: This follows because \( \left( {{W}_{1} + {W}_{2}}\right) /{W}_{1} \) and \( {W}_{2}/\left( {{W}_{1} \cap {W}_{2}}\right) \) are isomorphic. | Yes |
Lemma 3. Suppose \( V = {F}^{n}, * \) is nondegenerate, and \( W \subseteq V \) is a subspace.\na. \( \dim \left( W\right) + \dim \left( {W}^{ \bot }\right) = n \). | Proof: Let \( A \) be the matrix of \( * \) . For part a, let \( w \) denote \( \dim \left( W\right) \) ; choose a basis in which \( W \) has the first \( w \) standard unit vectors as a basis, and let \( {A}_{1} \) be the leftmost \( w \) columns of \( A \) . Since \( * \) is nondegenerate, the column space of \( {A}_... | Yes |
Theorem 6 (Sylvester’s law of inertia). Suppose \( V = {F}^{n};F \) is weakly \( \mathcal{R} \) -like and \( * \) is symmetric, or \( F \) is weakly \( \mathcal{C} \) -like and \( * \) is Hermitian; and \( * \) is nondegenerate. Then there is an integer \( r \) such that in any orthogonal basis \( S,\left| {\{ v \in S ... | Proof: Let \( {v}_{1},\ldots ,{v}_{n} \) be an orthogonal basis with \( {v}_{i} * {v}_{i} > 0 \) iff \( i \leq r \), and let \( {w}_{1},\ldots ,{w}_{n} \) be an orthogonal basis with \( {w}_{i} * {w}_{i} < 0 \) iff \( i \leq s \) . It suffices to show that \( {v}_{1},\ldots ,{v}_{r},{w}_{1},\ldots {w}_{s} \) are linear... | Yes |
Theorem 8 (Witt’s theorem). If \( {W}_{1},{W}_{2} \) are nondegenerate subspaces and \( T : {W}_{1} \mapsto {W}_{2} \) is an isomorphism then \( T \) can be extended to an automorphism of \( V \) . | Proof: Let \( u \in {W}_{1} \) be such that \( u * u \neq 0 \) . By lemma 7 there is an automorphism \( {T}_{0} \) such that \( {T}_{0}\left( u\right) = T\left( u\right) \) . Replacing \( T \) by \( {T}_{0}^{-1}T \) we may assume \( T\left( u\right) = u \) . If \( \dim \left( {W}_{1}\right) = 1 \) we are done; otherwis... | Yes |
Lemma 10. If \( * \) is \( r \) -nondegenerate then \( {\rho }_{r} \) is injective, and if \( * \) is \( r \) -regular then \( {\rho }_{r} \) is bijective. Similar claims hold for \( l \) . | Proof: \( \; \) Suppose \( x * T\left( y\right) = 0 \) for all \( x, y \) ; then for all \( {yT}\left( y\right) \in {K}_{r} \), so if \( * \) is r-nondegenerate then \( T\left( y\right) = 0 \) for all \( y \) . This proves that \( {\rho }_{r} \) is injective. Given any form \( x \circ y \), the map \( {\psi }_{y} = x \... | Yes |
Theorem 16. A Schur decomposition exists. | Proof: If \( n = 1 \) the claim is trivial. Otherwise let \( \lambda, x \) be such that \( {Ax} = {\lambda x} \) and \( \left| x\right| = 1 \) . Let \( S \) be any unitary matrix whose first column is \( x \) . One readily checks that\n\n\[ {S}^{ \dagger }{AS} = \left\lbrack \begin{array}{ll} \lambda & {A}_{1} \\ 0 & {... | Yes |
Lemma 2. Let \( F, G, H \) be formulas, \( x \) a variable, and \( t \) a term.\na. \( \vDash F \Rightarrow \left( {G \Rightarrow F}\right) \). | Proof: Exercise. | No |
Lemma 4. If \( S, H{ \vdash }_{\mathrm{{PC}}}G \) then \( S{ \vdash }_{\mathrm{{PC}}}H \Rightarrow G \) . | Proof: The proof is by induction on the length of the first derivation. If \( G = H \) the claim follows by lemma 3. If \( G \) is an axiom the claim follows since \( \vdash G \Rightarrow \left( {H \Rightarrow G}\right) \) . If \( G \) follows by rule e then inductively \( S \vdash H \Rightarrow F \) and \( S \vdash H ... | Yes |
Lemma 11. A consistent set \( T \) of formulas is a subset of a maximal consistent set \( U \) . Further, there is a maximal consistent set \( V \) in a language expanded with constants, where every formula \( \neg \forall {xF}\left( x\right) \in V \) has a witness in \( V \) . | Proof: The existence of \( U \) follows by the same argument as Lemma 5. Given a set \( S \) of formulas over some language, for each sentence \( \neg \forall {xF}\left( x\right) \) of the language add to \( S \) a sentence \( \neg \forall {xF}\left( x\right) \Rightarrow \neg F\left( c\right) \), where the constants \(... | Yes |
Lemma 15. Suppose \( B \) is a Boolean algebra. If \( I \) is an ideal then the relation \( x \equiv y \), which holds iff \( x \oplus y \in I \), is a congruence relation. Dually if \( F \) is a filter then the relation \( x \equiv y \), which holds iff \( x \oplus y \in F \) , is a congruence relation. | Proof: The relation \( \equiv \) is reflexive since \( x \oplus x = 0 \in I \) ; symmetric since \( \oplus \) is commutative; and transitive since \( x \oplus z = \left( {x \oplus z}\right) \oplus \left( {x \oplus z}\right) \) and an ideal is closed under \( \oplus \) . Suppose \( x \oplus {x}^{\prime }, y \oplus {y}^{... | No |
Corollary 17. In a distributive lattice \( L \), if \( a \nleqslant b \) there is a prime ideal (filter) \( H \) with \( b \in H \) and \( a \notin H \) \( \left( {a \in H\text{ and }b \notin H}\right) \) . | Proof: Apply the lemma with \( I = {b}^{ \leq } \) and \( F = {a}^{ \geq } \) . | No |
Lemma 8. Suppose \( n \geq 1 \) and \( x, y \geq 2 \). a. \( {A}_{n}\left( {x, y}\right) > y \). b. \( {A}_{n}\left( {x,\operatorname{Suc}\left( y\right) }\right) > {A}_{n}\left( {x, y}\right) \). c. \( {A}_{n}\left( {\operatorname{Suc}\left( x\right), y}\right) > {A}_{n}\left( {x, y}\right) \). d. \( {A}_{n + 1}\left(... | Proof: First, by induction on \( n{A}_{n}\left( {x,1}\right) = x \) for \( n \geq 2 \), and \( {A}_{n}\left( {2,2}\right) = 4 \) for \( n \geq 1 \). Parts a-c are verified directly for \( n = 1 \) ; suppose \( n \geq 2 \). For part a, the basis \( y = 2 \) has been verified; and \( {A}_{n}\left( {x, y}\right) > y \geq ... | Yes |
Lemma 9. Suppose \( n \geq 1 \) and \( x, y, z \geq 2 \) . a. \( {A}_{n}\left( {{A}_{n}\left( {x, y}\right), z}\right) \leq {A}_{n}\left( {x,{A}_{n}\left( {y, z}\right) }\right) \) . b. \( {A}_{n}\left( {{A}_{n + 1}\left( {x, y}\right) ,{A}_{n + 1}\left( {x, z}\right) }\right) \leq {A}_{n + 1}\left( {x, y + z}\right) \... | Proof: Parts a and b are proved for successive \( n \) ; equality holds everywhere for \( n = 1,2 \) for part a, and \( n = 1 \) for part b. Part a follows by induction on \( z \) using part b for \( n - 1 \) and the induction hypothesis. Part b follows by induction on \( y \) using part a and the induction hypothesis.... | Yes |
Corollary 20. If \( T \) is a \( {G}_{2} \) -theory then \( \neg ▱ \bot \) is not provable in \( T \) . | Proof: Apply Löb’s theorem with \( F \) being \( \bot \) . | Yes |
Lemma 25. Suppose \( L \) is a first order language and \( T \) a theory in \( L \). a. (Lowenheim-Skolem theorem) \( T \) has a model of cardinality \( \kappa \), where \( \kappa = \omega \) if \( L \) is finite, else \( \left| L\right| \) . | Proof: Part a follows by the construction in the proof of theorem 11.12; this holds for infinite languages, and the cardinality of the constructed model is as stated. | Yes |
Lemma 29. Suppose \( P\left( {{x}_{1},\ldots ,{x}_{k}}\right) \) is a predicate in \( \mathcal{P} \) and \( {P}_{n} \) are the Boolean functions of the preceding paragraph. There is a function \( {\operatorname{Rep}}_{P}\left( w\right) \) in \( \mathcal{L} \) which, on any input \( w \) of length \( n \) produces (the ... | Proof: Suppose \( P \) is computed by Turing machine \( M \) in time \( p\left( n\right) \) where \( n = \max \left( {\ell \left( {x}_{1}\right) ,\ldots ,\ell \left( {x}_{k}\right) }\right) \) and \( p \) is a suitable polynomial. The head is required to remain within a range of \( p\left( n\right) \) squares. The circ... | Yes |
Lemma 32. Suppose \( m \leq n \) when \( n - m \) is considered.\na. \( {x}_{n \pm m} = {x}_{n}{x}_{m} \pm d{y}_{n}{y}_{m},\;{y}_{n \pm m} = {x}_{m}{y}_{n} \pm {x}_{n}{y}_{m} \)\nb. \( {x}_{n \pm 1} = a{x}_{n} \pm d{y}_{n},\;{y}_{n \pm 1} = a{y}_{n} \pm {x}_{n} \)\nc. \( {x}_{n + 1} = {2a}{x}_{n} - {x}_{n - 1},\;{y}_{n... | Proof: Part a follows because\n\n\[ \n{x}_{n \pm m} + {y}_{n \pm m}\sqrt{d} = \left( {{x}_{n} + {y}_{n}\sqrt{d}}\right) \left( {{x}_{m} \pm {y}_{m}\sqrt{d}}\right) .\n\]\n\nPart \( \mathrm{b} \) is a special case, and part \( \mathrm{c} \) follows. | No |
Lemma 33. Suppose \( n \in \mathcal{N} \). a. \( \gcd \left( {{x}_{n},{y}_{n}}\right) = 1 \) b. \( {x}_{n + 1} > {x}_{n},\;{y}_{n + 1} > {y}_{n} > n,\;{a}^{n} \leq {x}_{n} \leq {\left( 2a\right) }^{n} \), and for \( n > 0\;{\left( 2a - 1\right) }^{n - 1} \leq {y}_{n} \leq {\left( 2a\right) }^{n - 1} \) c. \( n \) is ev... | Proof: Part a is obvious from Pell's equation. Part b follows by induction using lemma 32b and 32c. Part c follows by induction using part a and lemma 32c. For part \( \mathrm{d} \), by induction on \( k \) using lemma 32a, \( {y}_{n} \mid {y}_{nk} \) . Writing any \( t \) as \( {nk} + r \) where \( r < n,{y}_{t} = {x}... | Yes |
Lemma 34. The set \( \\left\\{ {\\left( {y, n, a}\\right) : y = {y}_{n}\\left( a\\right), n > {0a} > 1}\\right\\} \) is Diophantine. | Proof: Consider the system\n\n1. \( a > 1, n > 0 \)\n\n2. \( {x}^{2} - \\left( {{a}^{2} - 1}\\right) {y}^{2} = 1 \)\n\n3. \( {x}_{1}^{2} - \\left( {{a}^{2} - 1}\\right) {y}_{1}^{2} = 1 \)\n\n4. \( {x}_{2}^{2} - \\left( {{b}^{2} - 1}\\right) {y}_{2}^{2} = 1 \)\n\n5. \( b \\equiv 1{\\;\\operatorname{mod}\\;2}y \)\n\n6. \... | Yes |
Lemma 35.\na. \( {x}_{n}\left( b\right) - \left( {b - a}\right) {y}_{n}\left( b\right) \equiv {a}^{n}{\;\operatorname{mod}\;2}{ba} - {a}^{2} - 1 \)\nb. for \( n > 0 \) and \( a > 0 \), if \( b > {a}^{n} \) then \( {2ba} - {a}^{2} - 1 > {a}^{n} \) | Proof: Part a follows by induction on \( n \) by lemma 32c. Indeed, using the induction hypothesis and lemma \( {32}\mathrm{c},{x}_{n + 1}\left( b\right) - \left( {b - a}\right) {y}_{n - 1}\left( b\right) \equiv {2b}{a}^{n} - {a}^{n - 1}{\;\operatorname{mod}\;2}{ba} - {a}^{2} - 1 \) ; the right side is congruent to \( ... | Yes |
Lemma 38. The set \( \{ \left( {z, n, k}\right) : z = B\left( {n, k}\right) \} \) is Diophantine. | Proof: Consider the system\n\n1. \( u = {2}^{n} + 1 \)\n\n2. \( z \equiv \left\lfloor {{\left( u + 1\right) }^{n}/{u}^{k}}\right\rfloor {\;\operatorname{mod}\;u} \)\n\n3. \( z < u \)\n\nSuppose for some \( \left( {z, n, k}\right) \) the system has a solution. By 1 and lemma \( {37}\left\lfloor {{\left( u + 1\right) }^{... | Yes |
Lemma 39. Two integers \( x \) and \( y \) are disjoint iff \( B\left( {x + y, x}\right) \) is odd. | Proof: Suppose \( x \) has \( r \) 1’s, \( y \) has \( s \) 1’s and \( x + y \) has \( t \) 1’s. One verifies that \( t \leq r + s \), with equality holding iff \( x \) and \( y \) are disjoint. We claim that the highest power of 2 dividing \( x \) ! is \( x - r \), and the lemma follows. To prove the claim, suppose \(... | No |
Corollary 41. If disjointness is \( p \) -Diophantine then every predicate in \( \mathcal{{NP}} \) is \( p \) -Diophantine. | Proof: The auxiliary variables in the proof of the theorem may have polynomially bounded length if the Turing machine runs in polynomial time; the witness may also, by definition. Finally, the condition that \( T \) be a power of 2 may be expressed as \ | No |
a. A subcollection of a category determines a substructure iff the inclusion map is a functor. | Proof: The first three parts follow by universal algebra, and the last is readily verified directly. | No |
Theorem 6. The forgetful functor \( G : {\operatorname{Struct}}_{L} \mapsto \) Set determines limits, for any \( J \in \) Cat and any \( F \in {\operatorname{Struct}}_{L}^{J} \) . | Proof: First suppose there are no relation symbols. It suffices to consider the limit \( \langle c,\alpha \rangle \) of theorem 5 . Now, \( d \) must have \( c \) as its domain and \( {\beta }_{i} \) must equal \( {\alpha }_{i} \), so we must define a structure on \( c \) so that \( {\alpha }_{i} \) is a homomorphism. ... | No |
Corollary 7. Suppose \( T \) is a set of universally quantified Horn formulas. The forgetful functor \( G \) : \( {\operatorname{Mdl}}_{T}^{ + } \mapsto \) Set determines limits, for any \( J \in \) Cat and any \( F \in {\operatorname{Mdl}}_{T}^{+J} \) . | Proof: The structure \( c \) is a substructure of the product of the \( F\left( i\right) \) . Since the axioms are Horn formulas the product is in \( {\mathrm{{Mdl}}}_{T}^{ + } \), and so \( c \) is. | Yes |
Theorem 12. If nonempty products and equalizers exist in a category \( C \) then nonempty limits exist. Dually if nonempty coproducts and coequalizers exist then nonempty colimits exist. | Proof: Suppose \( F \in {C}^{J} \) is given. Let \( a \) be the product of the indexed family \( F\left( i\right), i \in J \), with cone \( \alpha \) ; and let \( b \) be the product of the indexed family \( F\left( {\operatorname{codom}\left( l\right) }\right), l \in {\operatorname{Ar}}_{J} \), with cone \( \beta \) .... | Yes |
Theorem 13. An equalizer is monic; dually a coequalizer is epic. | Proof: Suppose \( f : a \mapsto b \) is the equalizer of \( {h}_{1},{h}_{2} : b \mapsto c \), and \( f{g}_{1} = f{g}_{2} \) where \( {g}_{i} : d \mapsto a \) . Then \( {h}_{1}f{g}_{i} = {h}_{2}f{g}_{i} \) for \( i = 1,2 \), whence \( f{g}_{i} = {f\alpha } \) for a unique \( \alpha \), and \( {g}_{1} = {g}_{2} = \alpha ... | Yes |
Theorem 16. The functor \( {\operatorname{Hom}}^{a} \) from a category \( D \) to Set preserves limits and monics. Dually \( {\operatorname{Hom}}_{b} \) preserves limits and monics. | Proof: Suppose \( \langle d,\alpha \rangle \) is a limit of \( K \) in \( D \), and \( \langle c,\beta \rangle \) is any cone to \( {\operatorname{Hom}}^{a}K \) . For each \( x \in c{\beta }_{i}\left( x\right) \) is an arrow from \( a \) to \( K\left( i\right) \) ; further if \( K\left( t\right) : K\left( i\right) \map... | Yes |
Lemma 19. An Abelian category is balanced. | Proof: If \( f \) is monic then \( f \equiv \operatorname{Ker}\left( {\operatorname{Coker}\left( f\right) }\right) \) ; if \( f \) is also epic then \( \operatorname{Coker}\left( f\right) = 0 \), so \( f \equiv \operatorname{Ker}\left( 0\right) \equiv \iota \) and \( f \) is an isomorphism. | Yes |
Theorem 3. If a prime \( p \) divides \( \left| G\right| \) for a finite group \( G \) then \( G \) contains an element of order \( p \) . | Proof: This is vacuously true if \( \left| G\right| = 1 \) . If \( p \) divides \( \left| H\right| \) for any proper subgroup \( H \subset G \), inductively \( H \) contains an element of order \( p \), which is an element of order \( p \) in \( G \) . In the remaining case, the size of the conjugacy class of an elemen... | No |
Lemma 4. The double cosets \( {HxK}, x \in G \), form a partition of \( G \) . | Proof: Suppose \( {h}_{1}x{k}_{1} = {h}_{2}x{k}_{2} \) ; multiplying by \( H \) on left and \( K \) on right on the right \( H{x}_{1}K = H{x}_{2}K \) . | No |
Lemma 7. In a finite group, with \( H \) a subgroup, the following are equivalent: \( x \in N\left( H\right) ,{HxH} = {xH} \) , \( {HxH} = {Hx},\left| {HxH}\right| = \left| H\right| ,\left| H\right| = \left| {H \cap {x}^{-1}{Hx}}\right| . | Proof: \( x \in N\left( H\right) \) iff \( {Hx} = {xH} \) ; certainly if this is so then \( {HxH} = {xH} \), and conversely if \( {HxH} = {xH} \) then \( {Hx} \subseteq {xH} \), and since they are the same size they are equal. Similarly \( {Hx} = {xH} \) iff \( {HxH} = {Hx} \) . Clearly if \( {HxH} = {Hx} \) then \( \l... | Yes |
Lemma 13. For finite \( G \), if \( G \) is transitive and \( N \vartriangleleft G \) then the orbits of \( N \) are a system of imprimitivity for \( G \) . In particular \( N \) is half transitive, and if \( G \) is primitive and \( N \) nontrivial then \( N \) is transitive. | Proof: It suffices to observe that if \( {Nx} \) is an orbit of \( N \) then so is \( {gNx}, g \in G \), since \( {gNx} = {gN}{g}^{-1}{gx} = \) \( {Ngx} \) . | Yes |
Corollary 17. Suppose \( G \) is an \( m + 1 \) -transitive permutation group on \( S \) which has a regular normal subgroup \( N \) ; then the conclusions of the theorem hold. | Proof: \( {\operatorname{Stab}}_{x} \) is \( m \) -transitive on \( S - \{ x\} \) ; by the above mentioned correspondence, it acts as a group of automorphisms on \( N, m \) -transitively on \( N - \{ e\} \) . | No |
Lemma 12. If \( \gcd \left( {w, d}\right) = 1, d > 0 \), then for any \( n > 0 \) there is a \( t \) such that \( \gcd \left( {w + {td}, n}\right) = 1 \) . | Proof: Let \( \left\{ {p}_{i}\right\} \) be the prime divisors of \( n \) . It is readily seen that for each \( i \) there is a \( {t}_{i} \) such that \( {p}_{i} \nmid w + {t}_{i}d \) ; if \( {p}_{i} \mid d \) then \( {p}_{i} \nmid w \) so let \( {t}_{i} = 0 \), and if \( {p}_{i} \nmid d \) let \( {t}_{i} \) be any va... | Yes |
Lemma 14. If \( A = B \oplus C \) and \( B \subseteq D \subseteq A \) then \( D = B \oplus \left( {C \cap D}\right) \) . | Proof: Clearly \( B \oplus \left( {C \cap D}\right) \subseteq D \) . Suppose \( d \in D \) and \( d = b + c \) where \( b \in B, c \in C \) ; then \( c = d - b \in D \) . | No |
Corollary 16. Any submodule or quotient module of a semisimple module is semisimple. If \( M \) is the sum of simple submodules then \( M \) is semisimple. | Proof: A submodule is by lemma 14. A quotient module \( M/N \) is isomorphic to \( {N}^{\prime } \) where \( M = N \oplus {N}^{\prime } \) . A submodule of a sum of simple modules is a summand. | No |
Lemma 17 (Schur’s lemma). If \( L \) and \( M \) are simple \( R \) -modules and \( h : L \mapsto M \) is a nonzero homomorphism then \( h \) is an isomorphism. | Proof: The image of \( h \) must be \( M \) since it is not 0 ; the kernel must be 0 since it is not \( L \) . | Yes |
Lemma 18. Suppose \( M \) is a simple \( R \) -module.\na. If \( L \) is a simple left ideal of \( R \) and \( {LM} \neq 0 \) then \( M \) is isomorphic to \( L \).\nb. If \( x \in M \) is nonzero then \( {Rx} = M \). | Proof: For part a, If \( {rx} \neq 0, r \in L, x \in M \), the map \( s \mapsto {sx} \) is a nonzero module homomorphism from \( L \) to \( M \). For part \( \mathrm{b},{Rx} \subseteq M \) and \( {Rx} \neq 0 \) so \( {Rx} = M \) . | No |
Theorem 19. Suppose \( R \) is a semisimple ring. Every \( R \) -module is semisimple. There is a complete set of central idempotents \( \left\{ {e}_{i}\right\} \), such that each \( R{e}_{i} \) is a simple ring. Further every simple \( R \) -module \( M \) is isomorphic to a simple left ideal of exactly one \( R{e}_{i... | Proof: Any \( R \) -module is a quotient of \( { \oplus }_{i \in I}R \) for some \( I \), so is semisimple. Define \( {R}_{i}, i \in I \), to be the sum of the simple left ideals in an isomorphism class (that is, the collection of finite sums of elements from the ideals). By lemma 18 if \( {L}_{1},{L}_{2} \) are noniso... | Yes |
a. The expression of a character \( \chi \) as \( \mathop{\sum }\limits_{i}{a}_{i}{\chi }_{i}, i \in \mathcal{N} \), is unique. | Proof: Applying \( \chi \) to \( {e}_{i} \), we see that if \( \chi = \mathop{\sum }\limits_{i}{a}_{i}{\chi }_{i} \) then \( \chi \left( {e}_{i}\right) = {a}_{i}{\chi }_{i}\left( {e}_{i}\right) = {a}_{i}{n}_{i} \), so \( {a}_{i} = \chi \left( {e}_{i}\right) /{n}_{i} \). | No |
Lemma 1. Suppose \( A, B, C \) are subgroups of \( G \). a. \( {AB} \subseteq {BA} \) iff \( {AB} = A \sqcup B \) iff \( {AB} \) is a subgroup iff \( {AB} = {BA} \) . | Proof: Write \( a,{a}^{\prime } \) for members of \( A \), etc. For part a, if any \( {ba} \) can be rewritten as \( {a}^{\prime }{b}^{\prime } \) then any product \( {a}_{1}{b}_{1}{a}_{2}\cdots \) can be rewritten as an element of \( {AB} \) . Thus, if \( {BA} \subseteq {AB} \) then \( A \sqcup B \subseteq {AB} \), so... | No |
Lemma 2. Suppose \( a \vartriangleleft A, b \vartriangleleft B \) . Then\n\n\[ a \sqcup \left( {A \cap b}\right) \vartriangleleft a \sqcup \left( {A \cap B}\right) ,\;\left( {A \cap b}\right) \sqcup \left( {a \cap B}\right) \vartriangleleft A \cap B,\text{ and } \]\n\n\[ \frac{a \sqcup \left( {A \cap B}\right) }{a \sqc... | Proof: The first claim follows because \( a \) and \( A \cap B \) both normalize \( a, A \cap b \) . The second claim follows because \( a \cap B \) and \( A \cap b \) are both normal in \( A \cap B \) . Write \( \alpha \) for \( a \sqcup \left( {A \cap b}\right) \) and \( \beta \) for \( a \sqcup \left( {A \cap B}\rig... | Yes |
Corollary 4. If \( {A}_{n} \vartriangleleft \cdots \vartriangleleft {A}_{0} \) in the theorem is a composition series then \( m \leq n \) . | Proof: In the series from \( {A}_{i - 1} \) to \( {A}_{i} \), all but one of the quotient groups are trivial. In the series of all \( {B}_{ji} \) there are \( n \) nontrivial quotients, and \( m \leq n \) follows. | No |
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